Problem: A sheet of glass has a density of $2.5 \dfrac{\text{g}}{\text{cm}^3}$. What is the density of the glass in $\dfrac{\text{kg}}{\text{m}^3}$ ?
We will convert $2.5 \dfrac{\text{g}}{\text{cm}^3}$ to a density in $\dfrac{\text{kg}}{\text{m}^3}$ using the following conversion rates: There are $1000\text{ g}$ per $1\text{ kg}$. There are $100\,\text{cm}$ per $1\text{ m}$. Therefore, there are $(100\text{ cm})^3=1{,}000{,}000\text{ cm}^3$ per $1\text{ m}^3$. $\begin{aligned} &\phantom{=} \dfrac{2.5\,\text{g}}{1\text{ cm}^3} \cdot\dfrac{1\text{ kg}}{1000\text{ g}} \cdot \dfrac{1{,}000{,}000\text{ cm}^3}{1\text{ m}^3} \\\\ &=\dfrac{2.5 \cdot 1 \cdot 1{,}000{,}000\cdot \cancel{\text{g}} \cdot \text{kg} \cdot \cancel{\text{cm}^3}}{1 \cdot 1000 \cdot 1 \cdot \cancel{\text{cm}^3} \cdot \cancel{\text{g}} \cdot \text{m}^3} \\\\ &=\dfrac{2{,}500{,}000}{1000}\,\dfrac{\text{kg}}{\text{m}^3} \\\\ &=2500\,\dfrac{\text{kg}}{\text{m}^3} \end{aligned}$ In conclusion, the density of the glass in $\dfrac{\text{kg}}{\text{m}^3}$ is: $2500\,\dfrac{\text{kg}}{\text{m}^3}$